Anagram Check Techniques in DSA Python - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Sorting-Based Anagram Check

What is the output of this Python code that checks if two words are anagrams by sorting their letters?

DSA Python

def are_anagrams(s1, s2):
    return sorted(s1) == sorted(s2)

print(are_anagrams('listen', 'silent'))
print(are_anagrams('hello', 'bello'))

A False\nFalse

B True\nFalse

C True\nTrue

D False\nTrue

💡 Hint

Sorting the letters of both words and comparing them checks if they have the same letters.

The function sorts both strings and compares them. 'listen' and 'silent' have the same letters, so True. 'hello' and 'bello' differ, so False.

❓ Predict Output

intermediate

Output of Frequency Count Anagram Check

What is the output of this Python code that checks if two words are anagrams by counting letter frequencies?

DSA Python

from collections import Counter

def are_anagrams(s1, s2):
    return Counter(s1) == Counter(s2)

print(are_anagrams('triangle', 'integral'))
print(are_anagrams('apple', 'papel'))

A False\nTrue

B False\nFalse

C True\nFalse

D True\nTrue

💡 Hint

Counting how many times each letter appears helps check if two words are anagrams.

'triangle' and 'integral' have the same letter counts, so True. 'apple' and 'papel' also have the same counts, so True.

🔧 Debug

advanced

Find the Error in Anagram Check Code

This code tries to check if two strings are anagrams but has a bug. What error does it cause when run?

DSA Python

def are_anagrams(s1, s2):
    counts = {}
    for ch in s1:
        counts[ch] = counts.get(ch, 0) + 1
    for ch in s2:
        counts[ch] = counts.get(ch, 0) - 1
    return all(value == 0 for value in counts.values())

print(are_anagrams('abc', 'cab'))
print(are_anagrams('abc', 'abb'))

A True\nFalse

B False\nFalse

C True\nTrue

D KeyError

💡 Hint

The code uses a dictionary to count letters, adding for s1 and subtracting for s2.

The code correctly counts letters and checks if all counts return to zero. 'abc' and 'cab' are anagrams (True). 'abc' and 'abb' are not (False). No error occurs.

🧠 Conceptual

advanced

Which Method is Most Efficient for Large Strings?

For very long strings, which anagram check method is generally the most efficient in terms of time complexity?

A Sorting both strings and comparing them

B Using recursion to check all permutations

C Counting letter frequencies using a fixed-size array or dictionary

D Comparing each character of one string with every character of the other

💡 Hint

Counting frequencies can be done in one pass each, avoiding sorting overhead.

Counting frequencies is O(n) time, sorting is O(n log n), comparing each character pair is O(n^2), and recursion checking permutations is factorial time, so frequency counting is best.

❓ Predict Output

expert

Output of Case-Insensitive Anagram Check with Spaces

What is the output of this code that checks if two phrases are anagrams ignoring case and spaces?

DSA Python

def clean_string(s):
    return ''.join(ch.lower() for ch in s if ch.isalpha())

def are_anagrams(s1, s2):
    return sorted(clean_string(s1)) == sorted(clean_string(s2))

print(are_anagrams('Dormitory', 'Dirty room'))
print(are_anagrams('Conversation', 'Voices rant on'))

A True\nTrue

B False\nTrue

C True\nFalse

D False\nFalse

💡 Hint

Removing spaces and ignoring case helps check if phrases are anagrams.

Both phrases have the same letters ignoring spaces and case, so both comparisons return True.

Practice

1. What does it mean if two strings are anagrams of each other?

easy

A. They contain the same characters in any order

B. They have the same length but different characters

C. They start with the same character

D. They end with the same character

Solution

1. Step 1: Understand the definition of anagrams

   Anagrams are words or phrases made by rearranging the letters of another, so they must have the same characters.

2. Step 2: Compare the character sets

   If two strings have exactly the same characters, just in different orders, they are anagrams.

3. Final Answer:

   They contain the same characters in any order -> Option A

4. Quick Check:

   Anagrams = same characters [OK]

Hint: Check if sorted strings are equal to confirm anagrams [OK]

Common Mistakes:

• Thinking same length means anagrams
• Only checking first or last character
• Ignoring character counts

2. Which of the following Python code snippets correctly checks if two strings s1 and s2 are anagrams using sorting?

easy

A. s1 == s2[::-1]

B. sorted(s1) == sorted(s2)

C. s1.sort() == s2.sort()

D. len(s1) == len(s2)

Solution

1. Step 1: Understand sorting strings in Python

   The sorted() function returns a new sorted list of characters from the string.

2. Step 2: Compare sorted lists for equality

   If sorted(s1) equals sorted(s2), both strings have the same characters in the same counts.

3. Final Answer:

   sorted(s1) == sorted(s2) -> Option B

4. Quick Check:

   sorted strings equal means anagrams [OK]

Hint: Use sorted() to compare strings for anagram check [OK]

Common Mistakes:

• Using .sort() on strings (strings have no sort method)
• Comparing reversed strings instead of sorted
• Only checking length equality

3. What will be the output of the following code?

from collections import Counter
s1 = 'listen'
s2 = 'silent'
print(Counter(s1) == Counter(s2))

medium

A. False

B. TypeError

C. True

D. None

Solution

1. Step 1: Understand Counter usage

   Counter counts the frequency of each character in the string.

2. Step 2: Compare Counters for equality

   Since 'listen' and 'silent' have the same characters with same counts, their Counters are equal.

3. Final Answer:

   True -> Option C

4. Quick Check:

   Counter equality means anagrams [OK]

Hint: Use Counter to count chars and compare for anagrams [OK]

Common Mistakes:

• Expecting False because strings differ in order
• Confusing Counter with sorting
• Not importing collections module

4. Find the error in this anagram check code:

def are_anagrams(s1, s2):
    return s1.sort() == s2.sort()

medium

A. Missing import statement

B. Function should return True always

C. Comparison should be with reversed strings

D. Strings do not have a sort() method

Solution

1. Step 1: Check string methods

   Strings in Python do not have a sort() method; only lists do.

2. Step 2: Identify correct approach

   Use sorted(s1) and sorted(s2) which return sorted lists for comparison.

3. Final Answer:

   Strings do not have a sort() method -> Option D

4. Quick Check:

   sort() not valid on strings [OK]

Hint: Use sorted() on strings, not .sort() method [OK]

Common Mistakes:

• Trying to call .sort() on strings
• Comparing reversed strings instead of sorted
• Ignoring return value of sorted()

5. You want to check if two strings are anagrams but ignore spaces and capitalization. Which code snippet correctly implements this?

hard

A. sorted(s1.lower().replace(' ', '')) == sorted(s2.lower().replace(' ', ''))

B. Counter(s1) == Counter(s2)

C. s1 == s2

D. len(s1) == len(s2)

Solution

1. Step 1: Normalize strings by lowering case and removing spaces

   Use s1.lower().replace(' ', '') and similarly for s2 to ignore case and spaces.

2. Step 2: Sort normalized strings and compare

   Sorting ensures characters are in order for easy comparison.

3. Final Answer:

   sorted(s1.lower().replace(' ', '')) == sorted(s2.lower().replace(' ', '')) -> Option A

4. Quick Check:

   Normalize then sort to check anagrams ignoring spaces/case [OK]

Hint: Lowercase and remove spaces before sorting for anagram check [OK]

Common Mistakes:

• Not removing spaces before comparison
• Not converting to lowercase
• Using Counter without normalization